Question

Two capacitors of different capacitances are connected first

1. in series and then
2. in parallel across a dc source of 100 V.

If the total energy stored in the combination in the two cases are 40 mJ and 250 mJ respectively, find the capacitance of the capacitors.

Answer 1

For series combination,

`V = 1/2 (C_1C_2)/(C_1 + C_2) V^2`

⇒ `0.04 = 1/2 (C_1C_2)/(C_1 + C_2) xx (100)^2`   ...(i)

For parallel combination,

`u = 1/2 (C_1 + C_2)V^2`

⇒ `0.25 = 1/2 (C_1 + C_2) xx (100)^2`

or C₁ + C₂ = 0.5 × 10⁻⁴   ...(ii)

from eq (i), `0.04 = 1/2 xx (C_1C_2)/(0.5 xx 10^-4) xx (100)^2`

⇒ C₁C₂ = 0.04 × 10⁻⁸

Now (C₁ − C₂)² = (C₁ + C₂)² − 4C₁C₂

= (0.5 × 10⁻⁴)² − 4 × 0.04 × 10⁻⁸

= (0.25 − 0.16) × 10⁻⁸

= 0.09 × 10⁻⁸

C₁ − C₂ = 0.3 × 10⁻⁴   ...(iii)

On solving (ii) and (iii)

C₁ = 0.4 × 10⁻⁴ F and

C₂ = 0.1 × 10⁻⁴ F